 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

JEE Main 2015 (Offline)

When $5V$ potential difference is applied across a wire of length $0.1$ $m,$ the drift speed of electrons is $2.5 \times {10^{ - 4}}\,\,m{s^{ - 1}}.$ If the electron density in the wire is $8 \times {10^{28}}\,\,{m^{ - 3}},$ the resistivity of the material is close to :
A
$1.6 \times {10^{ - 6}}\Omega m$
B
$1.6 \times {10^{ - 5}}\Omega m$
C
$1.6 \times {10^{ - 8}}\Omega m$
D
$1.6 \times {10^{ - 7}}\Omega m$

Explanation

$V = IR = \left( {neA{v_d}} \right)\rho {\ell \over A}$

$\therefore$ $\rho = {V \over {{V_d}\ln e}}$

Here $V=$ potential difference

$l =$ length of wire

$n=$ no. of electrons per unit volume of conductor.

$e=$ no. of electrons

Placing the value of above parameters we get resistivity

$\rho = {5 \over {8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 4}} \times 0.1}}$

$= 1.6 \times {10^{ - 5}}\Omega m$
2

JEE Main 2014 (Offline)

In a large building, three are $15$ bulbs of $40$ $W$, $5$ bulbs of $100$ $W$, $5$ fans of $80$ $W$ and $1$ heater of $1$ $kW.$ The voltage of electric mains is $220$ $V.$ The minimum capacity of the main fuse of the building will be:
A
$8$ $A$
B
$10$ $A$
C
$12$ $A$
D
$14$ $A$

Explanation

Total power consumed by electrical appliances in the building, ${P_{total}} = 2500W$

Watt $=$ Volt $\times$ ampere

$\Rightarrow 2500 = V \times {\rm I}$

$\Rightarrow 2500 = 220$ ${\rm I}$

$\Rightarrow I = {{2500} \over {220}} = 11.36 \approx 12A$

(Minimum capacity of main fuse)
3

JEE Main 2013 (Offline)

This questions has Statement - ${\rm I}$ and Statement - ${\rm I}$${\rm I}. Of the four choices given after the Statements, choose the one that best describes into two Statements. Statement - {\rm I} : Higher the range, greater is the resistance of ammeter. Statement - {\rm I}$${\rm I}$ : To increase the range of ammeter, additional shunt needs to be used across it.

A
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is true, Statement - ${\rm II}$ is the correct explanation of statement - ${\rm I}$.
B
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is true, Statement - ${\rm II}$ is not the correct explanation of statement - ${\rm I}$.
C
Statement - ${\rm I}$ is true, Statement - ${\rm II}$ is false
D
Statement - ${\rm I}$ is false, Statement - ${\rm II}$ is true

Explanation

Statements ${\rm I}$ is false and Statement ${\rm I}$${\rm I}$ is true

For ammeter, shunt resistance, $S = {{{{\rm I}_g}G} \over {{\rm I} - {{\rm I}_g}}}$

Therefore for ${\rm I}$ to increase, $S$ should decrease, So additional $S$ can be connected across it.
4

JEE Main 2013 (Offline)

The supply voltage to room is $120V.$ The resistance of the lead wires is $6\Omega$. A $60$ $W$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $240$ $W$ heater is switched on in parallel to the bulb?
A
zero
B
$2.9$ Volt
C
$13.3$ Volt
D
$10.04$ Volt

Explanation Power of bulb $=60W$ $\left( {given} \right)$

Resistance of bulb $= {{120 \times 120} \over {60}} = 240\Omega$

$\left[ {\,\,} \right.$ $\left. {\,P = {{{V^2}} \over R}\,} \right]$

Power of heater $=240W$ (given)

Resistance of heater $= {{120 \times 120} \over {240}} = 60\Omega$

Voltage across bulb before heater is switched on,

${V_1} = {{240} \over {246}} \times 120 = 117.73\,\,$ volt

Voltage across bulb after heater is switched on,

${V_2} = {{48} \over {54}} \times 120 = 106.66$ volt

Hence decrease in voltage

${V_1} - {V_2} = 117.073 - 106.66 = 10.04$ Volt (approximately)